Faraday's Law of Induction — Question 7

Section 30-2 · Physics pages 4784–4785

Question 7 A changing long-wire current links a nearby loop logarithmically A fixed rectangular loop of height ell extends radially from a to b on one side of a long straight wire. The source current is I(t) = I0 cos ωt. Choose the loop normal along the field produced by positive I, and use B(r) = µ0I/(2πr). I(t) a b ℓ Schematic; labels, not drawing lengths, specify the data. (a) Find flux and emf. (b) Find the effect on emf amplitude of doubling both a and b while keeping ell fixed.
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Question 7 - Solution A changing long-wire current links a nearby loop logarithmically I(t) a b ℓ (a) Method and calculation Φ = ℓ ∫b a µ0I(t) 2πr dr = µ0ℓI0 2π ln(b/a) cos ωt. E = µ0ℓI0ω 2π ln(b/a) sin ωt . (b) Interpretation and further calculation ln 2b 2a = ln b a. E0,new = E0 . Physical check The outward shift weakens the field but doubles the radial width; these changes compensate for this scaled geometry.

Question 7

A changing long-wire current links a nearby loop logarithmically

A fixed rectangular loop of height ell extends radially from a to b on one side of a long straight wire. The source current is I(t)=I0cos⁡ωtI(t)=I_0\cos\omega t. Choose the loop normal along the field produced by positive I, and use B(r)=μ0I/(2πr)B(r)=\mu_0I/(2\pi r).

Diagram shown in the original SVG worksheet.

  1. Find flux and emf.

  2. Find the effect on emf amplitude of doubling both a and b while keeping ell fixed.

Question 7 - Solution

A changing long-wire current links a nearby loop logarithmically

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Φ=ℓ∫abμ0I(t)2πrdr=μ0ℓI02πln⁡(b/a)cos⁡ωt.\Phi=\ell\int_a^b\frac{\mu_0I(t)}{2\pi r}dr=\frac{\mu_0\ell I_0}{2\pi}\ln(b/a)\cos\omega t.ℰ=μ0ℓI0ω2πln⁡(b/a)sin⁡ωt.\boxed{\mathcal E=\frac{\mu_0\ell I_0\omega}{2\pi}\ln(b/a)\sin\omega t}.

(b) Interpretation and further calculation

ln⁡2b2a=ln⁡ba.\ln\frac{2b}{2a}=\ln\frac ba.ℰ0,new=ℰ0.\boxed{\mathcal E_{0,\mathrm{new}}=\mathcal E_0}.

Physical check

The outward shift weakens the field but doubles the radial width; these changes compensate for this scaled geometry.